Modern Physics 7 Question 46

51. A nucleus at rest undergoes a decay emitting an α-particle of de-Broglie wavelength, λ=5.76×1015m. If the mass of the daughter nucleus is 223.610amu and that of the α-particle is 4.002amu. Determine the total kinetic energy in the final state. Hence obtain the mass of the parent nucleus in amu.

(1amu=931.470MeV/c2)

(2001,5M)

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Answer:

Correct Answer: 51. 6.25×1011, zero, 5.0eV

Solution:

  1. (a) Given mass of α-particle, m=4.002 amu and mass of daughter nucleus,

M=223.610amu

de-Broglie wavelength of α-particle,

λ=5.76×1015m

So, momentum of α-particle would be

p=hλ=6.63×10345.76×1015kgm/s or p=1.151×1019kgm/s

From law of conservation of linear momentum, this should also be equal to the linear momentum of the daughter nucleus (in opposite direction).

Let K1 and K2 be the kinetic energies of α-particle and daughter nucleus. Then total kinetic energy in the final state is

K=K1+K2=p22m+p22M=p221m+1MK=p22M+mMm1amu=1.67×1027kg

Substituting the values, we get

or

K=1012JK=10121.6×1013=6.25MeV

(b) Mass defect, Δm=6.25931.470=0.0067amu

Therefore, mass of parent nucleus = mass of α-particle + mass of daughter nucleus + mass defect(Δm)

=(4.002+223.610+0.0067)amu=227.62amu

Hence, mass of parent nucleus is 227.62amu.



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